A) \[{{\tan }^{-1}}(4\pi )\]
B) \[{{\tan }^{-1}}(2\pi )\]
C) \[{{\tan }^{-1}}(1\pi )\]
D) \[{{\tan }^{-1}}(3\pi )\]
Correct Answer: B
Solution :
\[\tan \,\phi =\frac{\sqrt{{{Z}^{2}}-{{R}^{2}}}}{R}\] \[\begin{align} & =\frac{\sqrt{{{R}^{2}}+{{(2\pi nL)}^{2}}-{{R}^{2}}}}{R} \\ & =\frac{2\pi nL}{R} \\ & =\frac{2\times \pi \times 50\times 0.4}{20}=2\pi \\ \end{align}\] \[\phi ={{\tan }^{-1}}(2\pi )\]
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